Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
template <typename T>
ostream &operator+(ostream &out, const vector<T> &vec) {
  for (const auto &x : vec) {
    out << x << " ";
  }
  out << "\n";
  return out;
}
template <typename T>
ostream &operator*(ostream &out, const vector<T> &vec) {
  for (const auto &x : vec) {
    out + x;
  }
  return out;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &vec) {
  for (auto &x : vec) {
    in >> x;
  }
  return in;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> seq(n, vector<int>(m));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> seq[i][j];
      if (j) {
        if (seq[i][j] - seq[i][j - 1] != seq[0][j] - seq[0][j - 1]) {
          cout << "-1\n";
          return;
        }
      }
    }
  }
  int ans = 1e9;
  int b_0 = -1;
  auto solve = [&](int b0) {
    int b = b0, curr = b0;
    for (int i = 1; i < m; i++) {
      b += seq[0][i] - seq[0][i - 1];
      curr += b;
      if (b < 0) {
        return;
      }
    }
    for (int i = 0; i < n; i++) {
      if (b0 > seq[i][0]) {
        return;
      }
      curr += seq[i][0] - b0;
    }
    if (curr < ans) {
      ans = curr;
      b_0 = b0;
    }
  };
  for (int i = 0; i <= 500; i++) {
    solve(i);
  }
  vector<int> row(n), col(m);
  col[0] = b_0;
  ll fin = 0;
  for (int i = 1; i < m; i++) {
    col[i] = col[i - 1] + seq[0][i] - seq[0][i - 1];
  }
  for (int i = 0; i < n; i++) {
    row[i] = seq[i][0] - b_0;
  }
  for (const auto &val : row) {
    fin += val;
  }
  for (const auto &val : col) {
    fin += val;
  }
  cout << fin << '\n';
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < row[i]; j++) {
      cout << "row " << i + 1 << "\n";
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < col[i]; j++) {
      cout << "col " << i + 1 << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}